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Linear read-once and related Boolean functions
∗Vadim Lozin† Igor Razgon‡ Viktor Zamaraev Elena Zamaraeva¶
Nikolai Yu. Zolotykhk
Abstract
It is known that a positive Boolean function f depending on n variables has at
least n+ 1 extremal points, i.e. minimal ones and maximal zeros. We show thatf
has exactly n+ 1extremal points if and only if it is linear read-once.
The class of linear read-once functions is known to be the intersection of the classes of read-once and threshold functions. Generalizing this result we show that the class of linear read-once functions is the intersection of read-once and Chow functions. We also nd the set of minimal read-once functions which are not linear read-once and the set of minimal threshold functions which are not linear read-once. In other words, we characterize the class of linear read-once functions by means of minimal forbidden subfunctions within the universe of read-once and the universe of threshold functions. Within the universe of threshold functions the importance of linear read-once func-tions is due to the fact that they attain the minimum value of the specication number, which isn+ 1for functions depending onnvariables. In 1995 Anthony et al.
conjec-tured that for all other threshold functions the specication number is strictly greater thann+ 1. We disprove this conjecture by exhibiting a threshold non-linear read-once
function depending onnvariables whose specication number isn+ 1.
Keywords: threshold function; read-once function; linear read-once function; nested canalyzing function; canalyzing function; Chow function
1 Introduction
Linear read-once functions constitute a remarkable subclass of several important classes of Boolean functions and appear in the literature frequently under various names. In theoret-ical computer science, they are known as nested [1] or linear read-once [5]. The latter term is justied by the fact that a linear read-once function is a read-once function admitting a Boolean formula that can be constructed inductively in a linear fashion. In [5], it was
∗This paper extends and strengthens the results presented at the 28th International Conference on
Algorithmic Learning Theory [13].
†Mathematics Institute, University of Warwick, UK. Email: V.Lozin@warwick.ac.uk
‡Department of Computer Science and Information Systems, Birkbeck University of London, UK. Email:
Igor@dcs.bbk.ac.uk
Department of Computer Science, Durham University, South Road, Durham, DH1 3LE, UK. Email: viktor.zamaraev@gmail.com
¶Mathematics Institute, University of Warwick, UK. Email: E.Zamaraeva@warwick.ac.uk
kInstitute of Information Technology, Mathematics and Mechanics, Lobachevsky State University of
also proved that linear read-once functions are equivalent to 1-decision lists. Later and independently this class was introduced in [11] under the name nested canalyzing functions, as a subclass of canalyzing functions, which appear to be important in biological applica-tions [10]. The signicance of nested canalyzing funcapplica-tions in the applicaapplica-tions motivated their further theoretical study [9, 12]. In particular, [9] establishes the equivalence between nested canalyzing and unate cascade functions, which have been studied in the design of logic circuits and binary decision diagrams. All mentioned terms refer to the same class of Boolean functions which we call linear read-once.
In [5], it was shown that the class of linear read-once functions is the intersection of the classes of read-once and threshold functions. Generalizing this result we show that the class of linear read-once functions is the intersection of read-once and Chow functions. We also nd the set of minimal read-once functions which are not linear read-once and the set of minimal threshold functions which are not linear read-once. In other words, we characterize the class of linear read-once functions by means of minimal forbidden subfunctions within the universe of read-once and the universe of threshold functions.
Within the universe of threshold functions the importance of linear read-once functions is due to the fact that they attain the minimum value of the specication number, i.e. of the number of Boolean points that uniquely specify a function in this universe. To study the range of values of specication number of threshold functions one can be restricted to positive threshold functions depending on all their variables, in which case the functions can be completely specied by their sets of extremal points, i.e. maximal zeros and minimal ones. In other words, the specication number of a positive threshold function is upper bounded by the number of its extremal points. For a linear read-once function, these numbers coincide and equaln+ 1. In 1995 Anthony et al. [1] conjectured that for all other threshold functions the specication number is strictly greater thann+ 1.
Our result about the minimal threshold non-linear read-once functions seems to support this conjecture, since all these functions have specication number 2n. One more result
proved in this paper, which can be viewed as a supporting argument for the conjecture, states that a positive function f depending on n variables has exactly n+ 1 extremal points if and only if it is linear read-once. Nevertheless, rather surprisingly, we show that the conjecture is not true by exhibiting a positive threshold non-linear read-once function depending onn variables whose specication number isn+ 1.
The organization of the paper is as follows. All preliminary information related to the topic of the paper, including denitions and notation, is presented in Section 2. Section 3 is devoted to the number of extremal points in positive functions. In Section 4 we show that the class of linear read-once functions is the intersection of the classes of read-once and Chow functions, and identify the set of minimal read-once functions which are not linear read-once. In Section 5, we characterize the class of linear read-once functions in terms of minimal threshold functions which are not linear read-once and give a counterexample to the conjecture of Anthony et al.
2 Preliminaries
Let B = {0,1}. For a Boolean n-dimensional hypercube Bn we dene sub-hypercube Bn(xi1 = α1, . . . , xik = αk) as the set of all points of B
equal to αj for every j = 1, . . . , k. For a point x ∈ Bn we denote by x the point in Bn
with(x)i = 1 if and only if(x)i = 0 for every i∈[n].
For a Boolean function f = f(x1, . . . , xn) on Bn, k∈ [n], and αk ∈ {0,1} we denote
byf|xk=αk the Boolean function onB
n−1 dened as follows:
f|xk=αk(x1, . . . , xk−1, xk+1, . . . , xn) =f(x1, . . . , xk−1, αk, xk+1, . . . , xn).
For i1, . . . , ik ∈ [n] and α1, . . . , αk ∈ {0,1} we denote by f|x1=α1,...,xk=αk the function
(f|x1=α1,...,xk−1=αk−1)|xk=αk. We say that f|x1=α1,...,xk=αk is the restriction of f to x1 = α1, . . . , xk=αk. We also say that a Boolean functiong is a restriction (or subfunction) of
a Boolean functionf ∈Bn if there existi1, . . . , ik∈[n] andα1, . . . , αk∈ {0,1} such that g=f|x1=α1,...,xk=αk.
A variablexk is called irrelevant forf if f|xk=1 ≡f|xk=0, i.e.,f|xk=1(x) =f|xk=0(x)for
every x∈Bn−1. Otherwise,xk is called relevant forf. If xk is irrelevant forf we also say
thatf does not depend on xk.
2.1 Positive functions and extremal points
By we denote a partial order over the set Bn, induced by inclusion in the power set
lattice of then-set. In other words, xy if (x)i= 1 implies(y)i = 1. In this case we will
say that x is below y. When xy and x6=y we will sometimes write x≺y.
Denition 1. A Boolean functionf is called positive (also known as positive monotone
or increasing) iff(x) = 1 and xy implyf(y) = 1.
For a positive Boolean function f, the set of its false points forms a down-set and the
set of its true points forms an up-set of the partially ordered set(Bn,). We denote by
Zf the set of maximal false points,
Uf the set of minimal true points.
We will refer to a point inZf as a maximal zero of f and to a point inUf as a minimal
one of f. A point will be called an extremal point of f if it is either a maximal zero or a
minimal one off. We denote by
r(f) the number of extremal points of f.
2.2 Threshold functions
Denition 2. A Boolean functionf on Bn is called a threshold function if there exist n
weights w1, . . . , wn∈Rand a threshold t∈Rsuch that, for all(x1, . . . , xn)∈Bn,
f(x1, . . . , xn) = 0 ⇐⇒ n
X
i=1
wixi≤t.
(and separating hyperplanes) representing a given threshold function, and if there exists an inequality with non-negative weights, thenf is a positive function.
Letk∈N, k≥2. A Boolean functionf onBnisk-summable if, for somer∈ {2, . . . , k},
there existr(not necessarily distinct) false points x1, . . . ,xrandr(not necessarily distinct)
true points y1, . . . ,yr such that
Pr
i=1xi =Pri=1yi (where the summation is over Rn). A
function is asummable if it is notk-summable for all k≥2.
Theorem 1. [6] A Boolean function is a threshold function if and only if it is asummable. It is known (see e.g. Theorem 9.3 in [4]) that the class of threshold functions is closed under taking restrictions, i.e. any restriction of a threshold function is again a threshold function.
2.3 Chow functions
An important class of Boolean functions was introduced in 1961 by Chow [2] and is known nowadays as Chow functions. This notion can be dened as follows.
Denition 3. The Chow parameters of a Boolean function f(x1, . . . , xn) are the n+ 1
integers(w1(f), w2(f), . . . , wn(f), w(f)), wherew(f)is the number of true points off and wi(f) is the number of true points of f where xi is also true. A Boolean function f is a
Chow function if no other function has the same Chow parameters asf.
The importance of the class of Chow functions is due to the fact that it contains all threshold functions, which was also shown by Chow in [2].
2.4 Read-once, linear read-once and canalyzing functions
Denition 4. A Boolean function f is called read-once if it can be represented by a
Boolean formula using the operations of conjunction, disjunction, and negation in which every variable appears at most once. We say that such a formula is a read-once formula forf.
Denition 5. A read-once function f is linear read-once (lro) if it is either a constant
function, or it can be represented by a nested formula dened recursively as follows: 1. both literalsx and xare nested formulas;
2. x∨t, x∧t,x∨t, x∧tare nested formulas, where x is a variable and tis a nested
formula that contains neither x, nor x.
It is not dicult to see that an lro functionf is positive if and only if a nested formula
representingf does not contain negations.
In [5], it has been shown that the class of lro functions is precisely the intersection of threshold and read-once functions.
Denition 6. A Boolean function f = f(x1, . . . , xn) is called canalyzing1 if there exists i∈[n]such thatf|xi=0 or f|xi=1 is a constant function.
It is easy to see that iff is a positive canalyzing function thenf|xi=0 ≡0 orf|xi=1≡1,
for somei∈[n].
2.5 Specifying sets and specication number
LetF be a class of Boolean functions of nvariables, and let f ∈ F.
Denition 7. A set of points S ⊆Bn is a specifying set for f inF if the only function
inF consistent with f on S is f itself. In this case we also say that S species f in the
classF. The minimal cardinality of a specifying set forf in F is called the specication
number off (inF) and denoted σF(f).
LetHn be the class of threshold Boolean functions of nvariables. It was shown in [8]
and later in [1] that the specication number of a threshold function of n variables is at
leastn+ 1.
Theorem 2. [8, 1] For any threshold Boolean functionf of nvariables σHn(f)≥n+ 1.
Also, in [1] it was shown that the lower bound is attained for lro functions.
Theorem 3. [1] For any lro functionf depending on all its nvariables, σHn(f) =n+ 1.
2.6 Essential points
In estimating the specication number of a threshold Boolean functionf ∈ Hn it is often
useful to consider essential points off dened as follows.
Denition 8. A point x is essential for f (with respect to class Hn), if there exists a
functiong∈ Hn such thatg(x)6=f(x) and g(y) =f(y) for every y∈Bn, y6=x.
Clearly, any specifying set for f must contain all essential points for f. It turns out
that the essential points alone are sucient to specifyf inHn [3]. Therefore, we have the
following well-known result.
Theorem 4. [3] The specication number σHn(f) of a function f ∈ Hn is equal to the
number of essential points off.
The following result is a restriction of Theorem 4 in [15] (proved for threshold functions of many-valued logic) to the case of Boolean threshold functions.
Theorem 5. [15] A zero of a threshold function f is essential if and only if there is
separating hyperplane containing it.
2.7 The number of essential points vs the number of extremal points
It was observed in [1] that in the study of specication number of threshold functions, one can be restricted to positive functions. To prove Theorem 3, the authors of [1] rst showed that for a positive threshold functionf depending on all its variables the set of extremal
points species f. Then they proved that for any positive lro function f of n relevant
variables the number of extremal points isn+ 1.
In addition to proving Theorem 3, the authors of [1] also conjectured that lro functions are the only functions with the specication numbern+ 1in the classHn.
Conjecture 1. [1] Iff ∈ Hnhas the specication numbern+ 1, thenf is linear read-once.
In the next section, we show that this conjecture becomes a true statement if we replace `specication number' by `number of extremal points'. Nonetheless, in spite of this result supporting the conjecture, we conclude the paper with a counterexample disproving it.
3 Positive functions and the number of extremal points
The main goal of this section is to prove the following theorem.
Theorem 6. Let f =f(x1, . . . , xn) be a positive function with k ≥ 0 relevant variables.
Then the number of extremal points of f is at least k+ 1. Moreover f has exactly k+ 1 extremal points if and only iff is lro.
We will prove Theorem 6 by induction onn. The statement is easily veriable forn= 1. Letn > 1 and assume that the theorem is true for functions of at most n−1 variables. In the rest of the section we prove the statement for n-variable functions. Our strategy
consists of three major steps. First, we prove the statement for canalyzing functions in Section 3.2. This case includes lro functions. Then, in Section 3.3, we prove the result for non-canalyzing functionsf such that for each variablexiboth restrictionsf|xi=0andf|xi=1
are canalyzing. Finally, in Section 3.4, we consider the case of non-canalyzing functions
f depending on a variable xi such that at least one of the restrictions f|xi=0 andf|xi=1 is
non-canalyzing. In Section 3.1, we introduce some terminology and prove a preliminary result.
3.1 A property of extremal points
We say that a maximal zero (resp. minimal one) y off(x1, . . . , xn)corresponds to a variable xi if (y)i = 0 (resp. (y)i = 1). It is not dicult to see that for any relevant variablexi,
there exists at least one minimal one and at least one maximal zero corresponding toxi.
We say that an extremal point off corresponds to a set S of variables if it corresponds to
at least one variable inS.
Lemma 1. For every setS of k relevant variables of a positive function f, there exist at
leastk+ 1extremal points corresponding to this set.
Proof. LetS be a minimal counterexample and letP be the set of extremal points
corre-sponding to the variables in S. Without loss of generality we assume that S consists of
we may also assume that |P|=k and for every proper subset S0 of S there exist at least |S0|+ 1extremal points corresponding to S0. This implies, by Hall's Theorem of distinct
representatives [7], that there exists a bijection betweenS and P mapping variable xi to
a point ai∈P corresponding tox i.
Let a be any maximal zero inP. We denote by b the point which coincides with a in all
coordinates beyond the rstk, and for each i∈ {1,2, . . . , k}we dene the i-th coordinate
of b to be1 if ai is a maximal zero, and to be 0 if ai is a minimal one.
Assume rst that f(b) = 0 and let c be any maximal zero above b (possibly b=c). If(c)1 = . . .= (c)k = 1, then a ≺c, contradicting that a is a maximal zero. Therefore,
(c)i = 0 for some 1 ≤ i ≤ k and hence c is a maximal zero corresponding to xi ∈ S.
Moreover, c is dierent from any maximal zero aj ∈ P because the j-th coordinate of
aj ∈P is 0, while thej-th coordinate of c is 1.
Suppose now thatf(b) = 1and let c be any minimal one below b (possibly b=c). If (c)1 = . . .= (c)k = 0, then c≺a, contradicting the positivity of f. Therefore, (c)i = 1
for some1≤i≤k and hence c is a minimal one corresponding toxi ∈S. Moreover, c is
dierent from any minimal one aj ∈P because the j-th coordinate of aj ∈ P is 1, while
thej-th coordinate of c is 0.
A contradiction in both cases shows that there is no counterexamples to the statement of the lemma.
3.2 Canalyzing functions
Lemma 2. Let f = f(x1, . . . , xn) be a positive canalyzing function with k ≥ 0 relevant
variables. Then the number of extremal points off is at leastk+ 1. Moreoverf has exactly k+ 1 extremal points if and only iff is lro.
Proof. The casek= 0 is trivial, and therefore we assume thatk≥1.
Let xi be a variable of f such that f|xi=0 ≡ 0 (the case f|xi=1 ≡ 1 is similar). Let f0 =f|xi=0 and f1 =f|xi=1. Clearly, xi is a relevant variable of f, otherwise f ≡0, that
is, k= 0. Since every relevant variable of f is relevant for at least one of the functionsf0
andf1, we conclude thatf1 hask−1 relevant variables.
The equivalencef0≡0 implies that for every extremal point(α1, . . . , αi−1, αi+1, . . . , αn)
off1, the corresponding point(α1, . . . , αi−1,1, αi+1, . . . , αn)is extremal forf. For the same
reason, there is only one extremal point of f with the i-th coordinate being equal to 0, namely, the point with all coordinates equal to 1, except for the i-th coordinate. Hence, r(f) =r(f1) + 1.
1. Iff1 is lro, then f is also lro, since f can be expressed as xi∧f1. By the induction
hypothesis r(f1) =k and thereforer(f) =k+ 1.
2. Iff1is not lro, thenf is also not lro, which is easy to see. By the induction hypothesis
r(f1)> kand therefore r(f)> k+ 1.
3.3 Non-canalyzing functions with canalyzing restrictions
In this section, we study non-canalyzing positive functions such that for each variable xi
First we remark that all variables of those functions are relevant. Indeed, if such a function has an irrelevant variable then the function is canalyzing.
Claim 1. Let f =f(x1, . . . , xn) be a positive non-canalyzing function such that for each
variable xi both restrictions f|xi=0 and f|xi=1 are canalyzing. Then all variables of f are
relevant.
Proof. Letxi be irrelevant, then f|xi=0 ≡ f|xi=1. But f|xi=0, f|xi=1 are canalyzing, hence
there exists p∈[n]such that f|xi=0, xp=0 ≡f|xi=1, xp=0 ≡0 or f|xi=0, xp=1 ≡f|xi=1, xp=1 ≡
1. In the former casef|xp=0 ≡0, in the latter casef|xp=1 ≡1. In any casef is canalyzing.
Contradiction.
Claim 2. Let f =f(x1, . . . , xn) be a positive non-canalyzing function such that for each
variablexi both restrictionsf|xi=0 and f|xi=1 are canalyzing. Then for eachi,
(a) there exists a maximal zero that contains0's in exactly two coordinates one of which is i,
(b) there exists a minimal one that contains1's in exactly two coordinates one of which is i.
Proof. Fix an i and denote f0 = f|xi=0, f1 = f|xi=1. Since f0 is canalyzing, there exists p∈[n]such that f0|xp=0 ≡0 orf0|xp=1 ≡1. We claim that the latter case is impossible.
Indeed, the positivity of f and f0|xp=1 ≡1 imply f1|xp=1 ≡ 1, and therefore f|xp=1 ≡1.
This contradicts the assumption that f is non-canalyzing. Thus, f0|xp=0 ≡ 0. Now we
claim that the Boolean point y with exactly two 0's in coordinates i and p is a maximal
zero. Indeed, if f in at least one of three points above y is 0, then, by positivity of f, f|xi=0= 0 or f|xp=0, which contradicts the assumption that f is non-canalyzing.
Similarly, one can show that f1|xr=1 ≡1 for some r ∈[n]implying that the Boolean
point with exactly two1's in coordinates iand r is a minimal one.
Claim 3. Let f =f(x1, . . . , xn) be a positive non-canalyzing function such that for each
variablexi both restrictions f|xi=0 and f|xi=1 are canalyzing. Then there is a minimal one
y of Hamming weight2such that y is a maximal zero, unless n= 4 in which case f has 6 extremal points.
Proof. Consider a graphG0(resp. G1) with vertex set[n]every edgeijof which represents
a maximal zero (resp. minimal one) that contains0's (resp. 1's) in exactly two coordinates
iand j. By Claim 2, every vertex in G0 is covered by an edge and every vertex in G1 is
covered by an edge. From this it follows in particular that each graphG0,G1 has at least
dn/2e edges.
In terms of the graphsG0andG1, the claim is equivalent to saying thatG0andG1have
a common edge. It is not dicult to see that forn≤3 the graphs G1 and G0 necessarily
have a common edge. Let us show that this is also the case forn≥5.
Assume thatG0 and G1 have no common edges, i.e. every edge ofG0 is a non-edge (a
pair of non-adjacent vertices) inG1. Let us prove that
(*) every edge ij ofG0 forms a vertex cover inG1, i.e. every edge ofG1 shares a vertex
Indeed, let ij be an edge of G0 and assume that G1 contains an edge pq such that p is
dierent from i, j and q is dierent from i, j. Then the minimal one corresponding to
the edge pq of G1 is below the maximal zero corresponding to the edge ij of G0. This
contradicts the positivity off and proves (*).
Consider an edgeij inG0. Sincen≥5, then G0 has at least 3 edges, hence from (*)
we get that at least one ofi, j covers at least two edges of G1, sayicovers ipand iq. Let
psbe an edge of G0 covering p. Ifs=6 q, thenps does not cover the edgeiq ofG1 which
contradicts to (*). Ifs=q, let t be any vertex dierent fromi, j, p, q. The vertex tmust
be covered by some edge tr in G1. If r is dierent from i, j thentr does not cover ij in
G0. If r is dierent from p, q then tr does not cover pq in G0. In both cases we get a
contradiction to (*), hence for n ≥ 5 the graphs G0 and G1 necessarily have a common
edge and hence the result follows in this case.
It remains to analyze the casen= 4. Up to renaming variables, the only possibility for
G0 and G1 to avoid a common edge is for G0 to have edges 12 and 34and for G1 to have
edges13and24. In other words,(0,0,1,1)and(1,1,0,0)are maximal zeros and(1,0,1,0) and (0,1,0,1) are minimal ones. By positivity, this completely denes the function f,
except for two points (0,1,1,0) and (1,0,0,1). However, regardless of the value of f in
these points, both of them are extremal and hencef has6extremal points.
Claim 4. Let f =f(x1, . . . , xn) be a positive non-canalyzing function such that for each
variable xi both restrictions f|xi=0 and f|xi=1 are canalyzing. Let y be a minimal one
of Hamming weight 2 such that y is a maximal zero. Denote the two coordinates of y containing 1's by iand s, and let f0 =f|xi=0 and f1 =f|xi=1.
(a) Variable xs is relevant for both functions f0 andf1.
(b) If a point a= (α1, . . . , αi−1, αi+1, . . . , αn)∈Bn−1 is an extremal point of fαi, αi ∈ {0,1}, then a0 = (α
1, . . . , αi−1, αi, αi+1, . . . , αn−1)∈Bn is an extremal point of f.
Proof. First, we note that since y is a minimal one, f1|xs=1 ≡ 1. Similarly, since y is a
maximal zero,f0|xs=0≡0.
To prove (a), suppose to the contrary thatf0 does not depend onxs. Then f0|xs=1 ≡ f0|xs=0 ≡ 0, and therefore f0 ≡ 0, which contradicts the assumption that f is
non-canalyzing. Similarly, one can show thatxs is relevant for f1.
Now we turn to (b) and prove the statement for αi = 1. For αi = 0 the arguments are
symmetric.
Assume rst that αs = 1. Since y is a minimal one, we have f1(b) = 1 for all
b= (β1, . . . , βi−1, βi+1, . . . , βn)withβs= 1. Due to the extremality ofa, all its components
besidesαs are zeros. It follows that a0 =y, which is a minimal one by assumption.
It remains to assume thatαs = 0. Let a be a maximal zero for the functionf1. If a0 is
not a maximal zero forf, then there is a00 a0 withf(a00) = 0. Since a00a0 andαi= 1,
thei-th component of a00 is1. By its removal, we obtain a zero off1 that is strictly above
a in contradiction to the minimality of the latter.
Let a be a minimal one for the functionf1. If a0 is not a minimal one forf, then there
is a00 ≺ a0 with f(a00) = 1. The i-th component of a00 must be 0, since otherwise by its
removal we obtain a one forf1 strictly below a. Also, the s-th component of a00 must be
0, since this component equals 0 in a. But then a00 y with f(a00) = 1 and f(y) = 0, a
Lemma 3. Letf =f(x1, . . . , xn) be a positive non-canalyzing function such that for each
variablexi both restrictionsf|xi=0 andf|xi=1 are canalyzing. Then the number of extremal
points off is at least n+ 2.
Proof. By Claim 3 we may assume that there is a minimal one y that contains 1's in exactly two coordinates, sayi and s, such that y is a maximal zero. Denote f0 = f|xi=0
andf1 =f|xi=1.
Let P, P0, and P1 be the sets of relevant variables of f, f0, and f1, respectively. By
Claim 1, P is the set of all variables. Since any relevant variable of f is relevant for at
least one of the functionsf0, f1 and, by Claim 4 (a), xs is a relevant variable of both of
them, we have
n=|P| ≤ |P0∪P1|+ 1 =|P0|+|P1| − |P0∩P1|+ 1≤ |P0|+|P1|.
By Lemma 2, r(f0)≥ |P0|+ 1,r(f1)≥ |P1|+ 1. Finally, by Claim 4 (b) the number r(f)
of extremal points off is at leastr(f0) +r(f1)≥ |P0|+|P1|+ 2≥n+ 2.
3.4 Non-canalyzing functions containing a non-canalyzing restriction
Due to Lemmas 2 and 3 it remains to show the bound for a positive non-canalyzing function
f =f(x1, . . . , xn) such that for somei∈[n]at least one off0 =f|xi=0 and f1 =f|xi=1 is
non-canalyzing. Letk be the number of relevant variables of f and let us prove that the
number of extremal points off is at least k+ 2. Consider two possible cases:
(a) xi is a irrelevant variable of f;
(b) xi is a relevant variable off.
In case (a) the functionf|xi=0≡f|xi=1 is non-canalyzing and has the same number of
extremal points and the same number of relevant variables asf. By induction, the number
of extremal points off is at leastk+ 2.
Now let us consider case (b). Assume without loss of generality that i = n, and let f0 =f|xn=0 and f1 =f|xn=1. We assume that f0 is non-canalyzing and prove that f has
at leastk+ 2extremal points, wherek is the number of relevant variables off. The case
whenf0 is canalyzing, butf1 is non-canalyzing is proved similarly.
Let us denote the number of relevant variables of f0 by m. Clearly, 1 ≤m ≤k−1.
Exactlyk−1−mof krelevant variables off are irrelevant for the function f0. Note that
these k−1−m variables are necessarily relevant for the function f1. By the induction
hypothesis, the numberr(f0) of extremal points off0 is at least m+ 2.
We introduce the following notation:
C0 the set of maximal zeros of f corresponding toxn;
P0 the set of all other maximal zeros of f, i.e.,P0=Zf \C0;
C1 the set of minimal ones of f corresponding to xn;
For a setA⊆Bnwe will denote byA∗ the restriction ofAto the rstn−1coordinates, i.e.,A∗={(α1, . . . , αn−1) |(α1, . . . , αn−1, αn)∈A for someαn∈ {0,1}}.
By denition, the number of extremal points of f is
r(f) =|C0|+|P1|+|C1|+|P0|=|C0∗|+|P ∗ 1|+|C
∗ 1|+|P
∗
0|. (1)
We want to express r(f) in terms of the number of extremal points of f0 and f1.
For this we need several observations. First, we observe that if (α1, . . . , αn−1, αn) is an
extremal point forf, the point (α1, . . . , αn−1) is extremal for fαn. Furthermore, we have
the following straightforward claim.
Claim 5. P1∗ is the set of minimal ones of f0 and P0∗ is the set of maximal zeros of f1.
In contrast to the minimal ones of f0, the set of maximal zeros of f0 in addition to
the points in C0∗ may contain extra points, which we denote by N0∗. In other words, Zf0 =C∗
0∪N0∗. Similarly, besidesC1∗, the set of minimal ones off1 may contain additional
points, which we denote byN1∗. That is,Uf1 =C∗
1 ∪N1∗.
Claim 6. The set N0∗ is a subset of the set P0∗ of maximal zeros of f1. The set N1∗ is a
subset of the setP1∗ of minimal ones off0.
Proof. We will prove the rst part of the statement, the second one is proved similarly. Suppose to the contrary that there exists a point a= (α1, . . . , αn−1)∈N0∗\P0∗, which is a
maximal zero forf0, but is not a maximal zero forf1. Notice thatf1(a) = 0, as otherwise
(α1, . . . , αn−1,0)would be a maximal zero forf, which is not the case, since a∈/ C0∗. Since
a is not a maximal zero for f1, there exists a maximal zero b ∈ Bn−1 for f1 such that
a ≺ b. But then we have f0(b) = 1 and f1(b) = 0, which contradicts the positivity of
functionf.
From Claim 5 we haver(f0) =|Zf0 ∪Uf0|=|C0∗|+|N0∗|+|P1∗|, which together with
(1) and Claim 6 imply
r(f) =|C0∗|+|P1∗|+|C1∗|+|P0∗|=|C0∗|+|P1∗|+|C1∗|+|N0∗|+|P0∗\N0∗|
=r(f0) +|C1∗|+|P0∗\N0∗|.
(2)
Using the induction hypothesis we conclude thatr(f)≥m+ 2 +|C∗
1|+|P0∗\N0∗|. To
derive the desired boundr(f)≥k+ 2, in the rest of this section we show thatC1∗∪P0∗\N0∗
contains at leastk−m points.
Claim 7. Letxi, i∈[n−1], be a relevant variable for f1, which is irrelevant forf0. Then
every maximal zero forf1 corresponding to xi belongs toP0∗\N0∗ and every minimal one
forf1 corresponding to xi belongs to C1∗.
Proof. Let x∈N0∗ and assume(x)i = 0. Then by changing in x thei-th coordinate from
0 to 1 we obtain a point x0 with f0(x0) = 1 6= f0(x), since x is a maximal zero for f0.
This contradicts the assumption thatxi is irrelevant forf0. Therefore, (x)i = 1and hence
no maximal zero forf1 corresponding to xi belongs toN0∗, i.e. every maximal zero for f1
corresponding toxi belongs toP0∗\N0∗
Similarly, one can show that no minimal one forf1 corresponding toxi belongs to N1∗,
Recall that there are exactlyk−1−mvariables that are relevant forf1 and irrelevant
forf0. Lemma 1 implies that there are at leastk−mextremal points forf1 corresponding
to these variables. By Claim 7, all these points belong to the set C1∗ ∪P0∗ \N0∗. This
conclusion establishes the main result of this section.
Lemma 4. Let f = f(x1, . . . , xn) be a positive non-canalyzing function with k relevant
variables such that for some i ∈ [n] at least one of the restrictions f0 =f|xi=0 and f1 = f|xi=1 is non-canalyzing. Then the number of extremal points of f is at leastk+ 2.
4 Chow and read-once functions
In this section, we look at the intersection of the classes of Chow and read-once functions and show that this is precisely the class of lro functions. Thus, our result generalizes a result from [5] showing that the class of lro functions is the intersection of the classes of read-once and threshold functions.
There are two read-once functions that play a crucial role in our characterization of read-once Chow functions:
g1 =g1(x, y, z, u) = (x∨y)∧(z∨u);
g2 =g2(x, y, z, u) = (x∧y)∨(z∧u).
Lemma 5. Functionsg1, g2 and all the functions obtained from them by negation of
vari-ables are not Chow.
Proof. Functiong1 is not Chow, because g1 is dierent from (x∨z)∧(y∨u) (e.g. they
have dierent values at the pointx= 1,y= 0,z= 1,u= 0), but both functions have the
same Chow parameters (6,6,6,6,9). In a similar way, one can show that neither g2 nor
any function obtained fromg1 or g2 by negation of variables is Chow.
The following lemma shows that the class of Chow functions is closed under taking restrictions.
Lemma 6. Iff(x1, . . . , xn) is a Chow function, then any restriction of f is also Chow.
Proof. Suppose to the contrary that f has a restriction which is not a Chow function,
namely,
g=g(xik+1, . . . , xin) :=fxi1=α1,...,xik=αk,
for some i1, . . . , in ∈ [n], α1, . . . , αk ∈ {0,1} and g is not a Chow function. Then there
exists a function g0 =g0(xik+1, . . . , xin) with the same Chow parameters as g. We dene
functionf0(x1, . . . , xn) as follows:
f0(x1, . . . , xn) =
(
f(x1, . . . , xn) if (xi1, . . . , xik)6= (α1, . . . , αk),
g0(xk+1, . . . , xn) if (xi1, . . . , xik) = (α1, . . . , αk).
Sincew(g) =w(g0), we conclude thatw(f) =w(f0). Similarly, for everyi∈ {ik+1, . . . , in},
the equalitywi(g) =wi(g0) implies that wi(f) =wi(f0). Consequently,f andf0 have the
Lemma 7. Any canalyzing read-once function f, which is not lro, has a non-constant
non-canalyzing read-once function as a restriction.
Proof. Let f be a minimum counterexample to the claim. Since f is canalyzing, there
existsα, β ∈ {0,1} such that f|xi=α ≡β. We assume that α =β = 1, i.e. f|xi=1 ≡1, in
which casef =xi∨f|xi=0 (the other cases are similar).
Clearly, f|xi=0 is read-once, since any restriction of a read-once function is read-once.
Also,f|xi=0 is not lro, since otherwisef is lro, and hencef|xi=0 is not a constant function.
Sincef is a counterexample,f|xi=0 is canalyzing and has no non-constant non-canalyzing
read-once restrictions. But then we have a contradiction to the minimality off.
Theorem 7. For a read-once function f the following statements are equivalent:
(1) f is an lro function;
(2) f is a Chow function;
(3) f does not have g1 or g2 or any function obtained from g1 or g2 by negation of
variables as a restriction.
Proof. It is known that all lro functions are threshold [5] and all threshold functions are Chow [2]. Therefore,(1) implies(2).
To prove that(2)implies(3), we observe that by Lemma 6 any restriction off is Chow.
This together with Lemma 5 imply the conclusion.
Finally, to prove that (3) implies (1), we assume that f is positive and show that if f is non-lro, then it has as a restriction at least one of the functions g1 and g2 (in the
case of a non-positive function, similar arguments show thatf contains as a restriction a
function obtained fromg1 org2 by possibly negating some variables). Also, without loss of
generality we assume thatf is non-canalyzing, otherwise we would consider a non-constant
non-canalyzing restriction off which is guaranteed by Lemma 7.
Since f is a read-once function, there exist read-once functions f1 and f2 such that
eitherf =f1∧f2 orf =f1∨f2 and the sets of relevant variables off1 andf2 are disjoint.
We let f = f1 ∨f2, since the other case can be proved similarly. Suppose, one of the
functionsf1 and f2, sayf1, does not contain a conjunction in its read-once formula. Then
for any relevant variable xi of f1 we have f|xi=1 ≡ 1, which contradicts the assumption
thatf is non-canalyzing. Hence, both f1 and f2 necessarily contain conjunctions in their
read-once formulas. This means that there exist i1, . . . , in ∈[n], α5, . . . , αn ∈ {0,1} such
that
f1|xi5=α5,...,xik=αk =xi1 ∧xi2, and
f2|xik+1=αk+1,...,xin=αn=xi3 ∧xi4,
where{xi5, . . . , xik}and {xik+1, . . . , xin}are the sets of relevant variables of the functions f1 and f2, respectively. Consequently
f|xi5=α5,...,xin=αn =
f1|xi5=α5,...,xik=αk∨f2|xik+1=αk+1,...,xin=αn =
(xi1 ∧xi2)∨(xi3∧xi4).
5 Threshold functions and specication number
In this section, we turn to threshold functions and characterize the class of lro functions within this universe by a setG of minimal functions which are not linear read-once
(Sec-tion 5.1). All func(Sec-tions inGdepending onnvariables have specication number2n, which
can be viewed as an argument supporting Conjecture 1. Nevertheless, in Section 5.2 we disprove the conjecture.
5.1 Minimal non-lro threshold functions
Forn≥3, denote bygn the function dened by its DNF
gn=x1x2∨x1x3∨. . .∨x1xn∨x2. . . xn.
Lemma 8. For anyn≥3, the functiongn is positive, non-lro, threshold function,
depend-ing on all its variables, and the specication number ofgn is2n.
Proof. Clearly,gn depends on all its variables. Furthermore,gn is positive, since its DNF
contains no negation of a variable. Also, it is easy to verify thatgn is not canalyzing, and
thereforeg is non-lro.
Now, we claim that the CNF of gn is
(x1∨x2)(x1∨x3). . .(x1∨xn)(x2∨x3∨ · · · ∨xn).
Indeed, the equivalence of the DNF and CNF can be directly checked by expanding the latter and applying the absorption law:
(x1∨x2)(x1∨x3). . .(x1∨xn)(x2∨x3∨ · · · ∨xn)
= (x1∨x2x3. . . xn)(x2∨x3∨ · · · ∨xn)
=x1x2∨x1x3∨ · · · ∨x1xn∨x2x3. . . xn.
From the DNF and the CNF ofgn we retrieve the minimal ones
x1 = (1,1,0, . . . ,0),
x2 = (1,0,1, . . . ,0),
. . . . xn−1 = (1,0,0, . . . ,1),
xn= (0,1,1, . . . ,1),
and maximal zeros ofgn
y1 = (0,0,1, . . . ,1),
y2 = (0,1,0, . . . ,1),
. . . . yn−1 = (0,1,1, . . . ,0),
yn= (1,0,0, . . . ,0),
respectively (see Theorems 1.26, 1.27 in [4]).
It is easy to check that all minimal ones x1,x2, . . . ,xn satisfy the equation
and all maximal zeros y1,y2, . . . ,yn satisfy the equation
(n−2)x1+x2+x3+· · ·+xn=n−2.
Hence (n−2)x1+x2+x3+· · ·+xn ≤ n−1 is a threshold inequality representing the
functiongn.
For any positive threshold function f, which depends on all its variables, the set of
its extremal points species f. Furthermore, every essential point of f must belong to
each specifying set. Therefore, all essential points of gn are extremal. On the other
hand, by Theorem 5, all extremal points of gn are essential and therefore, by Theorem 4, σHn(gn) = 2n.
It is not dicult to see that gn (n≥ 3) is a minimal threshold function which is not
lro, i.e. any restriction of gn (n ≥ 3) is an lro function. Moreover, the same is true for
any function obtained from gn (n ≥ 3) by negation of variables, since the negation of a
variable of a threshold function results in a threshold function. We denote the set of all these minimal functions byG and show in what follows that there are no other minimal
threshold functions which are not lro.
Theorem 8. A threshold functionf is lro if and only if it does not contain any function
fromG as a restriction.
Proof. Stetsenko proved in [14] that the set of all, up to renaming and negation of variables, minimal not read-once functions consists of the following functions:
gn(x1, . . . , xn) =x1(x2∨. . .∨xn)∨x2. . . xn (n≥3); h1n(x1, . . . , xn) =x1. . . xn∨x1. . . xn (n≥2); h2n(x1, . . . , xn) =x1(x2∨x3. . . xn)∨x2x3. . . xn (n≥3); h3(x1, . . . , x5) =x1(x3x4∨x5)∨x2(x3∨x4x5);
h4(x1, . . . , x4) =x1(x2∨x3)∨x3x4.
Let us show that all functions in this list, except for gn, are 2-summable, hence are not
threshold.
• For the function h1n we have:
h1n(1,0, . . . ,0) =h1n(0,1, . . . ,1) = 0, h1
n(0,0, . . . ,0) =h1n(1,1, . . . ,1) = 1
and
(1,0, . . . ,0) + (0,1, . . . ,1) = (0,0, . . . ,0) + (1,1, . . . ,1).
• For the function h2n we have:
h2n(1,0,0, . . . ,0) =h2n(0,1,1, . . . ,1) = 0, h2
n(0,1,0, . . . ,0) =h2n(1,0,1, . . . ,1) = 1
and
• For the function h3 we have:
h3(0,0,1,1,1) =h3(1,1,0,0,0) = 0, h3(0,1,1,0,0) =h3(1,0,0,1,1) = 1
and
(0,0,1,1,1) + (1,1,0,0,0) = (0,1,1,0,0) + (1,0,0,1,1).
• For h4 we have:
h4(1,0,0,1) =h4(0,1,1,0) = 0,
h4(1,1,0,0) =h4(0,0,1,1) = 1 and
(1,0,0,1) + (0,1,1,0) = (1,1,0,0) + (0,0,1,1).
Since the functionsh1
n, h2n, h3, h4 are not threshold,f does not contain any of them or any
function obtained from them by negation of variables as a restriction. If, additionally, f
does not contain any function fromG, then f is read-once and hence is lro. Iff contains
a function fromG as a restriction, thenf is not read-once and hence is not lro.
5.2 Non-lro threshold functions with minimum specication number
Theorem 9. For a natural number n≥4, letfn=f(x1, . . . , xn) be a function dened by
its DNF
x1x2∨x1x3∨ · · · ∨x1xn−1∨x2x3. . . xn.
Then fn is positive, non-lro, threshold function, depending on all its variables, and the
specication number offn isn+ 1.
Proof. Clearly,fn depends on all its variables, it is positive, not canalyzing, and therefore f is non-lro.
It is easy to verify that CNF of fnis
(x1∨x2)(x1∨x3). . .(x1∨xn)(x2∨x3∨ · · · ∨xn−1).
From the DNF and the CNF offn we retrieve the minimal ones
x1= (1,1,0, . . . ,0,0),
x2= (1,0,1, . . . ,0,0),
. . . . xn−2= (1,0,0, . . . ,1,0),
xn−1= (0,1,1, . . . ,1,1)
and maximal zeros offn
y1= (0,0,1, . . . ,1,1),
y2= (0,1,0, . . . ,1,1),
. . . . yn−2= (0,1,1, . . . ,0,1),
z1= (0,1,1, . . . ,1,0),
It is easy to check that all minimal ones x1,x2, . . . ,xn−1 belong to the hyperplane
(2n−5)x1+ 2(x2+x3+· · ·+xn−1) +xn= 2n−3,
the maximal zeros y1,y2, . . . ,yn−2 belong to the hyperplane
(2n−5)x1+ 2(x2+x3+· · ·+xn−1) +xn= 2n−5,
and the maximal zeros z1,z2 belong to the hyperplane
(2n−5)x1+ 2(x2+x3+· · ·+xn−1) +xn= 2n−4.
Hence, (2n−5)x1 + 2(x2 +x3 +· · ·+xn−1) +xn ≤ 2n−4 is a threshold inequality
representingfn.
As in the proof of Lemma 8 we conclude that every essential point offn is extremal.
However, the extremal points y1,y2, . . . ,yn−2 are not essential. To show this, suppose to
the contrary that there exists a threshold functiondi that diers fromfnonly in the point
yi,i∈[n−2], i.e.,di(yi) = 1anddi(x) =fn(x)for every x6=yi. Then xi+yi =z1+z2,
and hence di is 2-summable. Therefore, by Theorem 1, di is not a threshold function, a
contradiction.
Now Theorems 4 and 5 imply that all the remainingn+ 1extremal points x1, . . . ,xn−1,
z1,z2 are essential, and thereforeσHn(fn) =n+ 1.
6 Conclusion
In this paper we proved a number of results related to the class of linear read-once functions. We also showed the existence of positive threshold Boolean functions ofnvariables, which
are not linear read-once and for which the specication number is at its lowest bound,
n+ 1. This leaves open the problem of characterizing the set of all such functions. We observe that this set is not closed under taking restrictions. In particular, the functions described in Theorem 9 contain, as restrictions, the functions from the setG. This example
also shows that specication number is not monotone with respect to restrictions, i.e. by restricting a function we can increase the specication number. All these remarks suggest that the problem of characterizing the set of all threshold functions with minimum value of specication number is highly non-trivial.
Acknowledgment The authors are grateful to Dmitry Chistikov for the useful discus-sions on the topic and for drawing their attention to the work of V. Stetsenko [14].
The work of Vadim Lozin and Nikolai Yu. Zolotykh was supported by the Russian Science Foundation Grant No. 17-11-01336.
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